# Math Help - Showing existence of an operator

1. ## Showing existence of an operator

I've been given the following question:

Let $(\mathcal{H},(,))$ be a real Hilbert space with an orthonormal basis $\left\{e_n\right\}_{n=1}^{\infty}$. Let $F:[0,1]\to \mathcal{H}$ be continuous. Show that there exists a positive self-adjoint operator $T\in B(\mathcal{H})$ such that $(Tx,y) = \int_0^1(F(t),x)(F(t),y)dt$ for all $x,y\in \mathcal{H}$.

Show moreover that $T$ is compact.

(you may use that $\int_0^1\sum_{n=N+1}^{\infty}|(F(t),e_n)|^2dt\to 0$ as $N\to\infty$)

I don't quite understand what must be shown here exactly. Must I find such a operator T explicitly and then show it has the desired property. If so, how? I don't quite see where the hint is useful, I guess it's meant for showing compactness of T.

Any hint or push in the right direction is very much appreciated.

2. [This is not a solution, or even a hint, but it might give you a push in the right direction.]

To get an idea of what is going on here, start by looking at the very simplest situation, when the function F is constant, say $F(t) = z\in\mathcal{H}$. Then $(Tx,y) = (z,x)(z,y)$, so $Tx = (z,x)z$. This is a rank-one operator (you may perhaps have seen the notation $T = z\otimes z$ for such an operator).

Next, take a slightly more complicated situation, where $F(t) = z_j$ for $(j-1)/n\leqslant z_j. (Admittedly, that function F is not continuous, but the idea is to use it as a step function giving a Riemann approximation to a continuous function.) Then $(Tx,y) = \frac1n\sum_{j=1}^n(z_j,x)(z_j,y)$. So $Tx = \frac1n\sum_{j=1}^n(z_j,x)z_j$, or $T= \frac1n\sum_{j=1}^nz_j\otimes z_j$. This is again a finite-rank operator.

Now use the Riemann approximation idea. Given a continuous function $F:[0,1]\to\mathcal{H}$, let $z_j = F(j/n)\;(1\leqslant j\leqslant n)$, and try to show that as $n\to\infty$ the operators $\frac1n\sum_{j=1}^nz_j\otimes z_j$ defined as in the previous paragraph converge in norm to T. Then T will be a norm limit of finite-rank operators, hence compact.